Photon travel time quadratic equation in Weinberg p. 72

[Kostik]

Thanks a lot strangerep. The decomposition appears to be $A = VDV^{-1}$, not $VDV^T$, so it doesn't quite match up with our equation $g=b^T\eta b$. Of course, $A^{-1}=A^T$ for orthogonal matrices, but is $b$ orthogonal?*

It is interesing to note that the definition of a Lorentz transformation is $Λ^T\eta Λ=\eta$ so we can see that when $g=\eta$ then $b=Λ$, which does agree with the known fact that b is only defined up to a Lorentz transformation.

I'll try to tidy this up ... I think it still isn't clear enough to me ... but it does seem that Weinberg made a little bit of a jump here.

*Edit: Yes. Since g is symmetric, in fact $g=VDV^T$ where D is the diagonal matrix of eigenvalues of g, and V is the matrix of g's [orthogonal] eigenvectors. But our equation is $g=b^T\eta b$, so $b=V^T$. And again here it's obvious that b (or V) is valid up to a Lorentz transformation because $Λ^T\eta Λ=\eta$.

Again many thanks. Your last post really helped me crack this.

 

[strangerep]

The decomposition appears to be $A = VDV^{-1}$, not $VDV^T$, so it doesn't quite match up with our equation $g=b^T\eta b$. Of course, $A^{-1}=A^T$ for orthogonal matrices, but is $b$ orthogonal?*

As you note in your edit, $A = VDV^T$ is correct if $g$ is real symmetric. Additional discussion here.

[...] it does seem that Weinberg made a little bit of a jump here.

Yes -- that's an example of what I meant when I said that Weinberg tends to be more suitable as an advanced text. He requires rather more of his readers, and sometimes this isn't obvious -- as in the present case.

 

[TSny]

It's just an application of eigendecomposition, i.e.,$$A ~=~ V D V^T ~,$$where $D$ is diagonal (with entries being the eigenvalues of $A$), and the columns of $V$ are the eigenvectors of a (real, symmetric) matrix $A$. In our case: $$g ~=~ b \eta b^T ~,$$ and the "constructive" solution for $b$ is obtained by finding eigenvectors of $g$.

Note that in general, $~g ~=~ b \eta b^T ~$ is not an eigendecomposition of $g$. The diagonal elements of $\eta$ are not, in general, the eigenvalues of $g$ and the columns of $b$ are not, in general, eigenvectors of $g$ .

As an example, suppose $g = \left( \begin{array}{cc} 7 & 4 \\ 4 & 1 \\ \end{array} \right)$.

Then you can show that $g = b \eta b^T$ where $\eta = \left( \begin{array}{cc} -1 &0 \\ 0 & 1 \\ \end{array} \right)$ and $b = \left( \begin{array}{cc} 3 & 4 \\ 0 & 1 \\ \end{array} \right)$

But the eigendecomposition of $g$ is $g = VDV^T$ where $V = \left( \begin{array}{cc} 1/\sqrt{5} &2/\sqrt{5} \\ -2/\sqrt{5} & 1/\sqrt{5} \\ \end{array} \right)$ and $D = \left( \begin{array}{cc} -1 &0 \\ 0 & 9 \\ \end{array} \right)$

 

[strangerep]

Note that in general, $~g ~=~ b \eta b^T ~$ is not an eigendecomposition of $g$. [...]

Heh, I was wondering whether someone would mention that.

In GR, the equivalence principle motivates an assumption that we have a metric field with signature (-,+,+,+) at each point of spacetime. Indeed, Weinberg actually starts from $\eta$ and transforms to $g$ -- see p71. His $b$'s are then also expressed as coord transformation coefficients -- see eqn(3.2.13).

IOW, we're not really starting from an arbitrary $g$, but one with the right eigenvalues.

 

[TSny]

Strangerep, I agree. Weinberg's discussion starts from the equivalence principle. This implies that (3.2.14) must have a solution where the $b$'s are associated with the coordinate transformation as in (3.2.13).

Still, the eigenvalues of $g(X)$ need not be the diagonal elements of $\eta$.

But, as you say, you can't just start from any symmetric matrix $g$. For example, taking the determinant of each side of (3.2.14) requires that the determinant of $g$ be negative.

 

[vanhees71]

The EP just says that you can, at any point $x_0$ in spacetime always find local coordinates in which $g_{\mu \nu}(x_0)=\eta_{\mu \nu}$. This implies that the symmetric matrix at each point must have 1 negative and 3 positive eigenvalues (Sylvester's theorem).

 

[samalkhaiat]

Weinberg starts this discussion in the middle of p. 72 by saying "The values of the metric tensor $g_{\mu\nu}$ and the affine connection $Γ^\lambda_{\mu\nu}$ at a point X in an arbitrary coordinate system $x^\mu$ PROVIDE ENOUGH INFORMATION TO DETERMINE THE LOCALLY INERTIAL COORDINATES $\xi^\alpha$ in a neighborhood of X." *This* is what I am not yet convinced of.

Thanks again for any help.

Weinberg’s statement is a provable statement. Let $x = 0$ be the point in question, and define new coordinates $y^{\mu}$ by
$$x^{\mu} = A^{\mu}{}_{\alpha}y^{\alpha} - \frac{1}{4} A^{\mu}{}_{\tau}B^{\tau}{}_{\alpha \beta}y^{\alpha}y^{\beta} , \ \ \ \ \ (1)$$ where $A^{\mu}{}_{\alpha}$ and $B^{\tau}{}_{\alpha \beta} = B^{\tau}{}_{\beta \alpha}$ are constants to be determined from reasonable requirements. At $x = 0$, any rank-2 tensor $g_{\mu\nu}(x)$ will therefore have the following transformation law $$\bar{g}_{\alpha\beta}(0) = A^{\mu}{}_{\alpha} \ g_{\mu\nu}(0) \ A^{\nu}{}_{\beta} . \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ In matrix form, (2) reads $$\bar{g} = A^{T} \ g \ A . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$

If $g_{\mu\nu}(x)$ is a metric tensor on some differentiable manifold $M^{n}$ then: (I) it must be symmetric, i.e., $g_{\mu\nu}(x) = g_{\nu\mu}(x)$. And, more importantly: (II) In order to model our spacetime by an equivalent class of pairs $(M^{n} , g_{\mu\nu})$, $g_{\mu\nu}$ must be a non-degenerate metric of Lorentz signature. (bellow I will be using the mostly minus signature).
Now,
(I) $\Rightarrow \ \exists R \in O(n)$ such that $R^{T}gR = G$ is diagonal, and
(II) $\Rightarrow \ G = \mbox{diag}(\lambda_{0}^{2}, - \lambda_{1}^{2}, \cdots , - \lambda_{n-1}^{2})$, where $\lambda_{r} \neq 0 \ \forall r$.
So, we have
$$R^{T} \ g \ R = \mbox{diag}(\lambda_{0}^{2}, -\lambda_{1}^{2}, -\lambda_{2}^{2} ,\cdots , -\lambda_{n-1}^{2}) . \ \ \ \ \ (4)$$ Define the matrix $D = \mbox{diag}(1/\lambda_{0},1/\lambda_{1}, \cdots , 1/\lambda_{n-1})$, and sandwich (4) by $D$ to obtain
$$DR^{T} \ g \ RD = \mbox{diag}(1, -1, -1, \cdots , -1) .$$ Thus $$(RD)^{T} \ g \ (RD) = \eta . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)$$ However, for any $\Lambda \in SO(1,n-1)$ , we also have $$(RD \Lambda )^{T} \ g \ (RD \Lambda ) = \Lambda^{T} \eta \Lambda = \eta .$$ So, we can set $$A^{\mu}{}_{\alpha} = (RD)^{\mu}{}_{\alpha} \ \ \mbox{mod} \ \Lambda \in SO(1,n-1) . \ \ \ \ \ \ (6)$$ Using the choice (6) together with (5), equation (2) or (3) gives us our first desired result, that is $$\bar{g}_{\alpha \beta}(0) = A^{\mu}{}_{\alpha} \ g_{\mu\nu}(0) \ A^{\nu}{}_{\beta} = \eta_{\alpha\beta} . \ \ \ \ \ \ (7)$$
The fact that the matrix $(RD)$ determines $A$ up to Lorentz transformations can also be deduced from parameter-counting: there are $n^{2}$ parameters in $A$, $n$ parameters in $D$ and $\frac{1}{2}n(n-1)$ parameters in the $O(n)$ matrix $R$. This leaves us with the $n^{2} - [\frac{1}{2}n(n-1) + n] = \frac{1}{2}n(n-1)$ free parameters needed for a Lorentz transformation.
To complete the proof of Weinberg’s statement, we need to choose the constants $B^{\mu}{}_{\alpha\beta}$ so that, together with the choice (6) for $A^{\mu}{}_{\alpha}$, the system $y$, as defined in (1), becomes locally inertial system. This requires a bit of algebra which I will describe them for you. Differentiating the transformation law $$\bar{g}_{\alpha\beta}(y) = \frac{\partial x^{\mu}}{\partial y^{\alpha}} \frac{\partial x^{\nu}}{\partial y^{\beta}} g_{\mu\nu}(x) ,$$ with respect to $y^{\gamma}$, we get at $x = 0$
$$\frac{\partial \bar{g}_{\alpha\beta}}{\partial y^{\gamma}}(0) = T_{\alpha\beta\gamma}(0) - \frac{1}{2}\left(A^{\mu}{}_{\alpha}g_{\mu\nu}(0) A^{\nu}{}_{\tau}\right) B^{\tau}{}_{\beta\gamma} - \frac{1}{2} \left(A^{\mu}{}_{\tau}g_{\mu\nu}(0)A^{\nu}{}_{\beta}\right) B^{\tau}{}_{\alpha\gamma} , \ \ (8)$$
where we have defined the object $$T_{\alpha\beta\gamma}(0) \equiv A^{\mu}{}_{\alpha}A^{\nu}{}_{\beta}A^{\rho}{}_{\gamma} \frac{\partial g_{\mu\nu}}{\partial x^{\rho}}(0) . \ \ \ \ (9)$$
Using our first result (7), we find
$$\frac{\partial \bar{g}_{\alpha\beta}}{\partial y^{\gamma}}(0) = T_{\alpha\beta\gamma}(0) - \frac{1}{2} (B_{\alpha\beta\gamma} + B_{\beta\alpha\gamma}) . \ \ \ \ (8')$$
So, to complete the proof we need to solve the following equation for $B_{\alpha\beta\gamma}$
$$T_{\alpha\beta\gamma} = \frac{1}{2}\left(B_{\alpha\beta\gamma} + B_{\beta\alpha\gamma}\right) . \ \ \ (10)$$
To do this, write another 2 copy of (10) using the permutation $(\alpha\beta\gamma) \to (\gamma\alpha\beta) \to (\beta\gamma\alpha)$, then add the first two and subtract the third one. Then, due to $B_{\alpha\beta\gamma} = B_{\alpha\gamma\beta}$, you obtain the final result
$$B_{\alpha\beta\gamma} = T_{\alpha\beta\gamma}(0) + T_{\gamma\alpha\beta} (0) - T_{\beta\gamma\alpha}(0) ,$$ with $T(0)$ as defined by (9).
So, from the values of $g_{\mu\nu}(0)$ and $\partial g (0) \sim \Gamma (0)$ we were able to determine (up to Lorentz transformation) the constants $A$ and $B$ that are needed to make the $y$-system locally inertial.